Question

这是我的疑问：

SELECT
   `i`.`itemtype` AS `Item`,
   `p`.`name`
   SUM(`i`.`count`) AS `Count`
FROM `player_items` AS `i`
LEFT JOIN `players` AS `p` ON (`p`.`id` = `i`.`player_id`)
WHERE `i`.`itemtype` IN (2148, 2152, 2160)
   AND `i`.`player_id` = " . intval ( $main_Char->getId() ) . "
GROUP BY `i`.`itemtype`
LIMIT 0, 30;

是否有可能为其选择的每个项目多次计数？我的意思是WHERE我. itemtype IN (2148, 2152, 2160) 它确定了所有ID项的计数：2148, 2152, 2160我需要它来做同样的事情，但另外要增加每个项目的计数：2148 * 10，2152 * 1000，2160 * 10000和然后在Count行显示结果。

Answer 1

是的，您可以一次SUM多个列：

SELECT i.itemtype AS Item
   , p.name
   , SUM(CASE 
            WHEN itemtype = 2148 THEN i.count
            ELSE 0
         END) * 10 AS Count
   , SUM(CASE
            WHEN itemtype = 2152 THEN i.count
            ELSE 0
         END) * 1000 AS Count1
   , SUM(CASE
            WHEN itemtype = 2160 THEN i.count
            ELSE 0
         END) * 10000 AS Count2
FROM player_items AS i
LEFT JOIN players AS p ON i.player_id = p.id
WHERE i.itemtype IN (2148, 2152, 2160)
GROUP BY i.itemtype ASC
LIMIT 0, 30

要将这三列合并在一起，您可以：

SELECT t.itemType AS Item, t.Count, t.Count1, t.Count2
   , SUM(t.Count1, t.Count2, t.Count3) AS Total
FROM
(
   SELECT i.itemtype AS Item
      , p.name
      , SUM(CASE 
               WHEN itemtype = 2148 THEN i.count
               ELSE 0
            END) * 10 AS Count
      , SUM(CASE
               WHEN itemtype = 2152 THEN i.count
               ELSE 0
            END) * 1000 AS Count1
      , SUM(CASE
               WHEN itemtype = 2160 THEN i.count
               ELSE 0
            END) * 10000 AS Count2
   FROM player_items AS i
   LEFT JOIN players AS p ON i.player_id = p.id
   WHERE i.itemtype IN (2148, 2152, 2160)
   GROUP BY i.itemtype ASC
   LIMIT 0, 30
) AS t

或者您可以添加另一个CASE

SELECT i.itemtype AS Item
   , p.name
   , SUM(CASE 
            WHEN itemtype = 2148 THEN i.count
            ELSE 0
         END) * 10 AS Count
   , SUM(CASE
            WHEN itemtype = 2152 THEN i.count
            ELSE 0
         END) * 1000 AS Count1
   , SUM(CASE
            WHEN itemtype = 2160 THEN i.count
            ELSE 0
         END) * 10000 AS Count2
   , SUM(CASE
            WHEN itemtype = 2148 THEN (i.count * 10)
            WHEN itemtype = 2152 THEN (i.count * 1000)
            WHEN itemtype = 2160 THEN (i.count * 10000)
            ELSE 0
         END) AS Total
FROM player_items AS i
LEFT JOIN players AS p ON i.player_id = p.id
WHERE i.itemtype IN (2148, 2152, 2160)
GROUP BY i.itemtype ASC
LIMIT 0, 30

Answer 2

目前还不是很清楚你想要什么。也许这个：

SUM(i.`count`) * COUNT(*) AS `Count`

或者这个：

SUM(i.`count`) * i.itemtype AS `Count`

查询中许多项的乘数

2 个答案: